Problem: Factor completely. $28-7x^2=$
Answer: First, we take a common factor of $7$. $28-7x^2=7(4-x^2)$ Now, let's factor $4-x^2$. Both $4$ and $x^2$ are perfect squares, since $4=({2})^2$ and $x^2=({x})^2$. $4-x^2 = ({2})^2-({x})^2$ So we can use the difference of squares pattern to factor. ${a}^2 - {b}^2 =({a}+{b})({a}-{b})$ In this case, ${a}={2}$ and ${b}={x}$ : $({2})^2 - ({x})^2 =({2}+{x})({2}-{x})$ $\begin{aligned} 28-7x^2&=7(4-x^2) \\\\ &=7(2+x)(2-x) \end{aligned}$ In conclusion, the complete factorization is $7(2+x)(2-x)$ Remember that you can always check your factorization by expanding it.